According to Mathematica, $$\frac{\pi}{2}\sum_{n=0}^\infty \frac{(4n)!}{(2n)!(n!)^2}x^n = \frac{1}{\sqrt{1+8\sqrt{x}}}K\left(\frac{16\sqrt{x}}{1+8\sqrt{x}}\right)$$ and similarly, $$\frac{\pi}{2} \sum_{n=0}^\infty \frac{(4n)!}{(2n)!(n!)^2}(-1)^n x^n = \frac{1}{\left(1+64x\right)^{1/4}}K\left(\frac{-1+\sqrt{1+64x}}{2\sqrt{1+64x}}\right)$$ where $K$ is the complete elliptic integral of the first kind. How can one prove these two expressions?
A relevant formula seems to be the "Landen transformation" listed in the NIST website. However it's unclear to me how this can be used to obtain the above.
Define the function $$ f(x):=\sum_{n=0}^\infty \frac{(4n)!}{(2n)!(n!)^2}x^n = {}_2F_1\Big(\frac14,\frac34;1;64x\Big). \tag{1}$$ This is the generating function of OEIS sequence A000897. We want to prove that $$ \frac\pi2 f(x^2) = \frac1{\sqrt{1+8x}}K\Big(\frac{16x}{1+8x}\Big). \tag{2} $$ Note that we can use the identity
$$ K(x) = \frac\pi2{}_2F_1\Big(\frac12,\frac12;1;x\Big), \tag{3} $$ equation $(1)$, and omitting factors of $\,\frac\pi2\,$ to rewrite equation $(2)$ as $$ \,{}_2F_1\Big(\frac14,\frac34;1;64x^2\Big) =\frac1{\sqrt{1+8x}} {}_2F_1\Big(\frac12,\frac12;1;\frac{16x}{1+8x}\Big). \tag{4} $$ From Berndt, Bhargava, and Garvan "Ramanujan's Theories of Elliptic Functions to Alternative Bases" (PDF link), pertaining to Ramanujan's theory of signature 4 there is Theorem 9.1, $${}_2F_1\Big(\frac12,\frac12;1;\frac{2x}{1+x}\Big)=\sqrt{1+x}\,{}_2F_1\Big(\frac14,\frac34;1;x^2\Big). \tag{5}$$ This theorem with $\,x\,$ replaced by $\,8x\,$ is easily shown to be equivalent to equation $(4)$, and this proves equation $(2)$.
We also want to prove $$ \frac\pi2 f(-x) = \frac{1}{\sqrt[4]{1+64x}}K\Big(\frac{-1+\sqrt{1+64x}}{2\sqrt{1+64x}}\Big). \tag{6}$$ Using equation $(1)$, equation $(3)$, and omitting factors of $\,\frac\pi2\,$ rewrite equation $(6)$ as $$ {}_2F_1\Big(\frac14,\frac34;1;-64x\Big) = \frac{1}{\sqrt[4]{1+64x}} {}_2F_1\Big(\frac12,\frac12;1; \frac{-1+\sqrt{1+64x}}{2\sqrt{1+64x}}\Big). \tag{7}$$ Replace $\,64x\,$ with $\,x\,$ to get $$ {}_2F_1\Big(\frac14,\frac34;1;-x\Big) = \frac{1}{\sqrt[4]{1+x}} {}_2F_1\Big(\frac12,\frac12;1; \frac{-1+\sqrt{1+x}}{2\sqrt{1+x}}\Big). \tag{8}$$
I don't know yet how to prove this.